Part B: Harder Problems

Transcript

0:00Introduction

Hello everyone! Here is a puzzle. In a Zener regulator, the input wanders between fifteen and twenty volts, but the output must stay at ten volts. You must choose the series resistor. Make it too big, and the Zener gets too little current. Make it too small, and it gets too much. Which resistors are safe? Example two will answer it. This is Part B of the unit problem bank. Problems like these come up often in JEE, so let's begin.

0:31Idea 1: donors and acceptors together

So, idea one. Add both kinds of dopant to silicon: pentavalent donors, which give electrons, and trivalent acceptors, which make holes. Do we now get lots of electrons and holes? We do not, because the two dopants partly cancel, and the larger one wins. Capital N D, with a small capital D below, is the number of donor atoms per unit volume, and N A, with a small capital A, the number of acceptor atoms; n e is the number of free electrons per unit volume. If N D is greater than N A, as the greater-than sign says, n e is about N D minus N A, where the wavy sign means approximately. Okay? Now for the minority carriers. In equilibrium, at a given temperature, n e times n h equals n i squared, pure or doped. Here n h is the number of holes per unit volume, and n i is the intrinsic carrier concentration; the small two means squared. Right? Now the conductivity, sigma, this Greek letter, equals e times, in brackets, n e times mu e, plus n h times mu h. Here e is the electronic charge, and mu e and mu h, with the Greek letter mu, are the electron and hole mobilities. In n-type material the holes are so few that sigma is about e times n e times mu e. Clear? And for silicon at three hundred kelvin, the capital K, n i is one point five times ten to the power sixteen per cubic metre, the m to the power minus three, and mu e is zero point one three five square metres per volt per second, the unit on screen. Got it?

2:17Idea 2: one current feeds the load and the Zener

Now idea two. On the left of the figure, capital I is the current and capital V the voltage; the dark blue curve is forward bias. In reverse, the orange curve drops at minus V Z, where V Z, with a small capital Z below, is the Zener voltage. There, in breakdown, the voltage stays at V Z while the current changes. On the right, the Zener, the triangle with a bent bar, is in reverse bias, in parallel with the load, so the output is V Z, okay? Capital I, the orange arrow, is the current through R S, the series resistor, with a small capital S below. It equals V in, the input voltage, minus V Z, over R S, while the load, R L, takes I L, equal to V Z over R L, with me? At the top black dot, the current I splits in two. So the Zener current, I Z, is I minus I L: the Zener takes the rest. Right? Now, suppose V in rises a little, with the same load. What happens to the Zener current? Yes, exactly, the Zener current rises. The current I grows, the load still takes the same I L, so the extra goes into the Zener. The extra volts sit across R S, and the output stays at V Z, clear?

3:39Example 1: conductivity of Si with donors and acceptors

Now Example one. Silicon is doped with ten to the power twenty two donors and four times ten to the power twenty one acceptors per cubic metre. Find n e, n h and the conductivity, with the silicon values from idea one. Pause the video and try it first. Done? Okay, let's check. Donors give electrons and acceptors give holes. So n e and n h are just these two numbers, right? That is exactly the trap, because the two dopants partly cancel. N D is greater than N A, so the crystal is n-type. Ten to the power twenty two is ten times ten to the power twenty one, so n e is about ten minus four, which is six, times ten to the power twenty one per cubic metre. With me? Now the holes: n h is n i squared over n e, and n i squared is two point two five times ten to the power thirty two. Divide by six times ten to the power twenty one, and n h is three point seven five times ten to the power ten per cubic metre, far fewer than the acceptors. Okay? Now the conductivity. Since this is n-type, sigma is about e times n e times mu e. That is six times ten to the power twenty one, times one point six times ten to the power minus nineteen, times zero point one three five. Right? Six times one point six is nine point six. The powers of ten leave a hundred, so we get nine hundred and sixty. Multiply by zero point one three five to get one hundred and twenty nine point six, about one hundred and thirty siemens per metre, the capital S m to the power minus one. Clear?

5:29Example 2: series resistor range for a 10 V Zener

Now Example two, our puzzle from the start. The input of a Zener regulator varies between fifteen and twenty volts. The Zener voltage, V Z, is ten volts, and the load takes a steady twenty milliamperes, the m A on screen. The Zener current must stay between five and sixty milliamperes. Find the allowed range of R S, the series resistor. Pause the video and try it first. Done? Okay, let's check. The output stays at ten volts, but the Zener current changes with the input. At which input is it smallest? Yes, exactly, at the lowest input, fifteen volts, where the current I is smallest, while the load takes the same. Got it? Now, at fifteen volts the Zener needs at least five milliamperes. So the current I through R S must be at least five milliamperes, right? Be careful, because that is the trap. The load also takes its twenty milliamperes from this same current. So I must be at least twenty plus five, which gives twenty five milliamperes; this sign means greater than or equal to. With me? And I is V in minus V Z, over R S, so a bigger R S gives a smaller current. So R S is at most fifteen minus ten, over zero point zero two five ampere, which gives two hundred ohms. This sign means less than or equal to, and the Greek capital omega means ohm. Okay? Now the other end. The Zener current is largest at the highest input, twenty volts, and must not go above sixty milliamperes. So I is at most twenty plus sixty, which gives eighty milliamperes. Right? So R S is at least twenty minus ten, over zero point zero eight ampere, which gives one hundred and twenty five ohms. Any smaller, and the Zener current goes above sixty milliamperes. Clear? So R S can be from one hundred and twenty five to two hundred ohms, the safe window from our puzzle. The lowest input fixes the largest resistor, and the highest input the smallest. Got it?

7:43Quick revision: ND minus NA, mass action law, Zener currents

So, quick revision. With both dopants, the larger one wins: n e is about N D minus N A, then n h is n i squared over n e. In n-type material, sigma is about e times n e times mu e. And in a Zener regulator, one current feeds the load and the Zener, so check the lowest and the highest input. Now try the other problems on this page. All the best!